If x is an integer and (x^2 - 2)!/(x^2-2)=33!, what is the rang

Question

If x is an integer and  \frac{(x^2 - 2)!}{x^2-2}=33! , what is the range of all possible values of x ?

A. 0
B. 2
C. 6
D. 11
E. 12

Bunuel · Accepted Answer

Official Solution:

If  x  is an integer and  \frac{(x^2 - 2)!}{x^2-2} = 33! , what is the range of all possible values of  x ?

A. 0
B. 2
C. 6
D. 11
E. 12

Since  \frac{n!}{n} = \frac{(n-1)!*n}{n}=(n-1)! , then:

\frac{(x^2 - 2)!}{x^2-2} =

=\frac{(x^2 - 2-1)!*(x^2 - 2)}{x^2-2} =

=(x^2 - 2-1)!=

=(x^2 - 3)!

Thus, we'd have:

(x^2 - 3)!=33!

x^2 - 3=33

x^2=36

x=-6  or  x=6

Thus, the range of all possible values of  x  is (the largest value) - (the smallest value)  = 6 - (-6) = 12 .

Answer: E

gmatophobia · Answer

We can have 34 in the denominator, so that equation transforms to

\frac{34! }{ 34} = 33 !

x^2 - 2  = 34

x^2 = 36

x = +6/ -6

Set of possible values of x = {-6,6}

Range = 12

Option E

SaquibHGMATWhiz · Answer

Solution:

We are given  \frac{(x^2 - 2)!}{x^2-2}=33! 
 ⇒\frac{(x^2-2)	imes (x^2-3)	imes (x^2-4)....	imes 2	imes 1}{x^2-2}=33! 
 ⇒(x^2-3)	imes (x^2-4)	imes (x^2-5)....	imes 2	imes 1=33! 
 ⇒(x^2-3)!=33!  
 We can compare both sides and say  x^2-3=33 
 ⇒x^2=36 
 ⇒x=+6 and -6  
 Range  =6-(-6)=6+6=12

Hence the right answer is    Option E

Flypaper3533 · Answer

Does the fact that x cannot take values such as  0 ,  1  or  \sqrt{2}  not impact the range? IMO the range is shorter because  |x|  must be > \sqrt{2}

Bunuel · Answer

The range is  the largest element  -  the smallest element , so the answer to your question is no.

samc88 · Answer

Set x^2 - 2 to be  n   then you get

n!/n = (n-1)!

Sub the value back in for n

(x^2 - 2) - 1 = 33!

x^2 - 3 = 33!

X = -6 OR 6

Range = 12

suman1031 · Answer

The range is from -6 to 6 means x can be 0, 1 , 2 or any value between -6 and 6.

Then how the equation is satisfied with the value x= 2, 3 or any value less than 6 ?

Bunuel · Answer

Only two values of x satisfy  \frac{(x^2 - 2)!}{x^2-2}=33! , namely -6 and 6. The range is (the largest element) - (the smallest element), so the range of all possible values of x is 6 - (-6) = 12.

Hope it's clear.

Archit3110 · Answer

value of LHS
 \frac{(x^2 - 2)!}{x^2-2}
(x^2-2) * ( x^2-3)! / (x^2-2)

( x^2-3)!
RHS is 33!
x can be +/-6
range will be 6-(-6) ; 12
option E

vishalgaurav4444 · Answer

n ! / n = (n -1)!
So (x2−2)!x2−2 = (x2−2 - 1)! 
So (x2−3)! = 33!
So x2−3 = 33
So x2 = 36
So x = +6 or -6
Range = 12

Option E

mysterymanrog · Answer

(x^2-2)!=(x^2-2)(x^2-3)! 
Knowing the above we can rewrite:
 \frac{(x^2-2)(x^2-3)!}{(x^2-2)}=33! 
 (x^2-3)!=33! 
X is either 6 or -6. The range is therefore 12.

laflem · Answer

(x^2−3)!=33!

On this step, how can we simplify to (x^2-3)=33? Conceptually, I am lost since factorials do not work that way.

Any explanation would be great.

Thanks

Bunuel · Answer

Factorials of different non-negative integers are all unique, except for 0! = 1! = 1. So if (x^2 - 3)! = 33!, then x^2 - 3 must equal 33. That’s why we can directly set x^2 - 3 = 33 and solve.

If x is an integer and (x^2 - 2)!/(x^2-2)=33!, what is the rang

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GMAT Problem Solving (PS) Questions

If x is an integer and (x^2 - 2)!/(x^2-2)=33!, what is the rang

Prep Toolkit

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